Here are summary statistics for randomly selected weights of newborn girls: nequals161 , x overbarequals32.8 hg, sequals7.2 hg. construct a confidence interval estimate of the mean. use a 90 % confidence level. are these results very different from the confidence interval 31.7 hgless thanmuless than34.5 hg with only 12 sample values, x overbarequals33.1 hg, and sequals2.7 hg? what is the confidence interval for the population mean mu ? 31.6 hgless thanmuless than 34.6 hg (round to one decimal place as needed.)

We have:

n = 161

xbar = 32.8hg

s = 7.2hg

cl = 90% = 0.90

Since we have only the sample standard deviation s, we need to use the t-distribution.

The degrees of freedom DF = 161 - 1 = 160

α = (1 - 0.90) / 2 = 0.05

If you look at these values in a t-distribution table you find t = 1.65

Now we can build the confidence interval:

xbar + / - (t · s / √n) = 32.8 + / - (1.65 · 7.2 / √161)

Therefore:

31.86 < μ < 33.74

In order to understand if these values are different from the ones you are given, let's calculate the confidence interval for the latest:

n = 12

xbar = 33.1hg

s = 2.7hg

31.7 < μ < 34.5

Calculate the error: (34.5 - 33.1) = (33.1 - 31.7) = 1.4

We know t · s / √n = 1.4

and we can solve for t:

t = 1.4 · √n / s = 1.4 · √12 / 2.7 = 1.7962

Looking at a t-distribution table, we find α = 0.05

which brings to a confidence level of 90%, which is the same for the previous part.

Since the sample size is big enough, we can use the normal distribution and the z-score:

Looking at a normal distribution table, we find z-score = 1.645, which is very similar to the t-value found previously. We don't know the population standard deviation, but for such a big sample the sample standard deviation is a good estimate, therefore:

xbar + / - (z * · s / √n) = 32.8 + / - (1.645 · 7.2 / √161)

31.8 < μ < 33.7